MHT CET · Maths · Application of Derivatives
The area of the square increases at the rate of \(0.5 \mathrm{~cm}^{2} / \mathrm{sec}\). The rate at which its
perimeter is increasing when the side of the square is \(10 \mathrm{~cm}\) long, is
- A \(0 \cdot 3 \mathrm{~cm} / \mathrm{sec}\)
- B \(0 \cdot 1 \mathrm{~cm} / \mathrm{sec}\)
- C \(0 \cdot 2 \mathrm{~cm} / \mathrm{sec}\)
- D \(0 \cdot 4 \mathrm{~cm} / \mathrm{sec}\)
Answer & Solution
Correct Answer
(B) \(0 \cdot 1 \mathrm{~cm} / \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
(D)
Let \(x\) be the side of the square. Then area \(A=x^{2}\)
\(\therefore \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}} \Rightarrow 0.5=2(10) \frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\)
\(\therefore \frac{\mathrm{dx}}{\mathrm{dt}}=0.025\)
Now perimeter \(=4 \mathrm{x}=\mathrm{p}\)
\(\therefore \frac{\mathrm{dp}}{\mathrm{dt}}=4 \frac{\mathrm{dx}}{\mathrm{dt}}=4(0.025)=0.1 \mathrm{~cm} / \mathrm{sec} .\)
Let \(x\) be the side of the square. Then area \(A=x^{2}\)
\(\therefore \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}} \Rightarrow 0.5=2(10) \frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\)
\(\therefore \frac{\mathrm{dx}}{\mathrm{dt}}=0.025\)
Now perimeter \(=4 \mathrm{x}=\mathrm{p}\)
\(\therefore \frac{\mathrm{dp}}{\mathrm{dt}}=4 \frac{\mathrm{dx}}{\mathrm{dt}}=4(0.025)=0.1 \mathrm{~cm} / \mathrm{sec} .\)
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